Prove that the largest isosceles triangle that can be drawn in a circle, is an equilateral triangle?

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Prove that the largest isosceles triangle that can be drawn in a circle, is an equilateral triangle?

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Answer 1 · 2017-03-19T15:21:05

Please see below.

Explanation:

Let their be an isosceles triangle ABC inscribed in a circle as shown, in which equal sides $AC$ and $BC$ subtend an angle $x$ at the center. It is apparent that side $AB$ subtends an angle $360^0-x$ at the center (as shown). Note that for equilateral triangles all these angles will be $(2pi)/3$ .

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As the area of the triangle portion subtended by an angle $x$ is $R^2/2sinx$ ,

the complete area of triangle ABC is

$A=R^2/2(sinx+sinx+sin(360-2x)$

= $R^2/2(2sinx-sin2x)$

= $R^2(sinx-sinxcosx)$

= $R^2sinx(1-cosx)$

For maximization we should have $(dA)/(dx)=0$

i.e. $R^2(cosx(1-cosx)+sinx xx sinx)=0$

or $cosx-cos^2x+1-cos^2x=0$

or $2cos^2x-cosx-1=0$

or $2cos^2x-2cosx+cosx-1=0$

or $2cosx(cosx-1)+1(cosx-1)=0$

or $(2cosx+1)(cosx-1)=0$

Hence $cosx=-1/2$ or $cosx=1$

i.e. $x=(2pi)/3$ or $x=0$

But for a triangle $x!=0$

hence $x=(2pi)/3$

and hence for maximum area triangle must be equilateral.

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Prove that the largest isosceles triangle that can be drawn in a circle, is an equilateral triangle?

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Explanation:

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